3.5.32 \(\int \frac {(a+b \text {ArcSin}(\frac {\sqrt {1-c x}}{\sqrt {1+c x}}))^3}{1-c^2 x^2} \, dx\) [432]

Optimal. Leaf size=275 \[ \frac {i \left (a+b \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^4}{4 b c}-\frac {\left (a+b \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \log \left (1-e^{2 i \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {3 i b \left (a+b \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {PolyLog}\left (2,e^{2 i \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c}-\frac {3 b^2 \left (a+b \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {PolyLog}\left (3,e^{2 i \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c}-\frac {3 i b^3 \text {PolyLog}\left (4,e^{2 i \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{4 c} \]

[Out]

1/4*I*(a+b*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^4/b/c-(a+b*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3*ln(1-(I*(-
c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))^2)/c+3/2*I*b*(a+b*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2)))
^2*polylog(2,(I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))^2)/c-3/2*b^2*(a+b*arcsin((-c*x+1)^(1/
2)/(c*x+1)^(1/2)))*polylog(3,(I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))^2)/c-3/4*I*b^3*polylo
g(4,(I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))^2)/c

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Rubi [A]
time = 0.16, antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6813, 4721, 3798, 2221, 2611, 6744, 2320, 6724} \begin {gather*} -\frac {3 b^2 \text {Li}_3\left (e^{2 i \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{2 c}+\frac {3 i b \text {Li}_2\left (e^{2 i \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{2 c}+\frac {i \left (a+b \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^4}{4 b c}-\frac {\log \left (1-e^{2 i \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3}{c}-\frac {3 i b^3 \text {Li}_4\left (e^{2 i \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3/(1 - c^2*x^2),x]

[Out]

((I/4)*(a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^4)/(b*c) - ((a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3*L
og[1 - E^((2*I)*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])])/c + (((3*I)/2)*b*(a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c
*x]])^2*PolyLog[2, E^((2*I)*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])])/c - (3*b^2*(a + b*ArcSin[Sqrt[1 - c*x]/Sqrt
[1 + c*x]])*PolyLog[3, E^((2*I)*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])])/(2*c) - (((3*I)/4)*b^3*PolyLog[4, E^((2
*I)*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])])/c

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3798

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[I*((c + d*x)^(m + 1)/(d*(
m + 1))), x] - Dist[2*I, Int[(c + d*x)^m*E^(2*I*k*Pi)*(E^(2*I*(e + f*x))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x))))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4721

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n*Cot[x], x], x, ArcSin[c*
x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 6744

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a + b*x)))^p]/(b*c*p*Log[F])), x] - Dist[f*(m/(b*c*p*Log[F])), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rule 6813

Int[((a_.) + (b_.)*(F_)[((c_.)*Sqrt[(d_.) + (e_.)*(x_)])/Sqrt[(f_.) + (g_.)*(x_)]])^(n_.)/((A_.) + (C_.)*(x_)^
2), x_Symbol] :> Dist[2*e*(g/(C*(e*f - d*g))), Subst[Int[(a + b*F[c*x])^n/x, x], x, Sqrt[d + e*x]/Sqrt[f + g*x
]], x] /; FreeQ[{a, b, c, d, e, f, g, A, C, F}, x] && EqQ[C*d*f - A*e*g, 0] && EqQ[e*f + d*g, 0] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx &=-\frac {\text {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^3}{x} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}\\ &=-\frac {\text {Subst}\left (\int (a+b x)^3 \cot (x) \, dx,x,\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )}{c}\\ &=\frac {i \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^4}{4 b c}+\frac {(2 i) \text {Subst}\left (\int \frac {e^{2 i x} (a+b x)^3}{1-e^{2 i x}} \, dx,x,\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )}{c}\\ &=\frac {i \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^4}{4 b c}-\frac {\left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \log \left (1-e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {(3 b) \text {Subst}\left (\int (a+b x)^2 \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )}{c}\\ &=\frac {i \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^4}{4 b c}-\frac {\left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \log \left (1-e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {3 i b \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {Li}_2\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c}-\frac {\left (3 i b^2\right ) \text {Subst}\left (\int (a+b x) \text {Li}_2\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )}{c}\\ &=\frac {i \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^4}{4 b c}-\frac {\left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \log \left (1-e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {3 i b \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {Li}_2\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c}-\frac {3 b^2 \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_3\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \text {Li}_3\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )}{2 c}\\ &=\frac {i \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^4}{4 b c}-\frac {\left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \log \left (1-e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {3 i b \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {Li}_2\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c}-\frac {3 b^2 \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_3\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c}-\frac {\left (3 i b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{4 c}\\ &=\frac {i \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^4}{4 b c}-\frac {\left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \log \left (1-e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {3 i b \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {Li}_2\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c}-\frac {3 b^2 \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_3\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c}-\frac {3 i b^3 \text {Li}_4\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{4 c}\\ \end {align*}

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Mathematica [F]
time = 0.21, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a+b \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3/(1 - c^2*x^2),x]

[Out]

Integrate[(a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3/(1 - c^2*x^2), x]

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1231 vs. \(2 (300 ) = 600\).
time = 0.89, size = 1232, normalized size = 4.48

method	result	size
default	\(-\frac {a^{3} \ln \left (c x -1\right )}{2 c}+\frac {a^{3} \ln \left (c x +1\right )}{2 c}-\frac {6 i b^{3} \polylog \left (4, \frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}-\frac {b^{3} \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{3} \ln \left (1-\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}+\frac {i b^{3} \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{4}}{4 c}-\frac {6 b^{3} \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \polylog \left (3, \frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}+\frac {i a \,b^{2} \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{3}}{c}-\frac {b^{3} \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{3} \ln \left (\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1-\frac {-c x +1}{c x +1}}+1\right )}{c}+\frac {6 i a \,b^{2} \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \polylog \left (2, -\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}-\frac {6 b^{3} \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \polylog \left (3, -\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}+\frac {3 i b^{3} \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2} \polylog \left (2, -\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}+\frac {3 i a^{2} b \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{2 c}-\frac {3 a \,b^{2} \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2} \ln \left (1-\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}-\frac {6 i b^{3} \polylog \left (4, -\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}-\frac {6 a \,b^{2} \polylog \left (3, \frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}-\frac {3 a \,b^{2} \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2} \ln \left (\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1-\frac {-c x +1}{c x +1}}+1\right )}{c}+\frac {3 i a^{2} b \polylog \left (2, \frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}-\frac {6 a \,b^{2} \polylog \left (3, -\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}+\frac {3 i b^{3} \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2} \polylog \left (2, \frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}-\frac {3 a^{2} b \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1-\frac {-c x +1}{c x +1}}+1\right )}{c}+\frac {3 i a^{2} b \polylog \left (2, -\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}-\frac {3 a^{2} b \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (1-\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}+\frac {6 i a \,b^{2} \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \polylog \left (2, \frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}\)	\(1232\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3/(-c^2*x^2+1),x,method=_RETURNVERBOSE)

[Out]

-1/2*a^3/c*ln(c*x-1)+1/2*a^3/c*ln(c*x+1)+6*I*a*b^2/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))*polylog(2,I*(-c*x+1)
^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))-b^3/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))^3*ln(1-I*(-c*x+1)^
(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))+1/4*I*b^3/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))^4-6*b^3/c*arc
sin((-c*x+1)^(1/2)/(c*x+1)^(1/2))*polylog(3,I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))+3*I*b^3
/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*polylog(2,-I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2)
)-b^3/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))^3*ln(I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2)+1)
-6*I*b^3/c*polylog(4,-I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))-6*b^3/c*arcsin((-c*x+1)^(1/2)
/(c*x+1)^(1/2))*polylog(3,-I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))+I*a*b^2/c*arcsin((-c*x+1
)^(1/2)/(c*x+1)^(1/2))^3+6*I*a*b^2/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))*polylog(2,-I*(-c*x+1)^(1/2)/(c*x+1)^
(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))-3*a*b^2/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*ln(1-I*(-c*x+1)^(1/2)/(c*x+
1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))+3/2*I*a^2*b/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2-6*a*b^2/c*polylog(3,
I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))-3*a*b^2/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*ln
(I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2)+1)+3*I*a^2*b/c*polylog(2,-I*(-c*x+1)^(1/2)/(c*x+1)^
(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))-6*a*b^2/c*polylog(3,-I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/
2))-6*I*b^3/c*polylog(4,I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))-3*a^2*b/c*arcsin((-c*x+1)^(
1/2)/(c*x+1)^(1/2))*ln(I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2)+1)+3*I*b^3/c*arcsin((-c*x+1)^
(1/2)/(c*x+1)^(1/2))^2*polylog(2,I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))-3*a^2*b/c*arcsin((
-c*x+1)^(1/2)/(c*x+1)^(1/2))*ln(1-I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))+3*I*a^2*b/c*polyl
og(2,I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3/(-c^2*x^2+1),x, algorithm="maxima")

[Out]

1/2*a^3*(log(c*x + 1)/c - log(c*x - 1)/c) - integrate((b^3*arctan2(sqrt(-c*x + 1), sqrt(2)*sqrt(c)*sqrt(x))^3
+ 3*a*b^2*arctan2(sqrt(-c*x + 1), sqrt(2)*sqrt(c)*sqrt(x))^2 + 3*a^2*b*arctan2(sqrt(-c*x + 1), sqrt(2)*sqrt(c)
*sqrt(x)))/(c^2*x^2 - 1), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3/(-c^2*x^2+1),x, algorithm="fricas")

[Out]

integral(-(b^3*arcsin(sqrt(-c*x + 1)/sqrt(c*x + 1))^3 + 3*a*b^2*arcsin(sqrt(-c*x + 1)/sqrt(c*x + 1))^2 + 3*a^2
*b*arcsin(sqrt(-c*x + 1)/sqrt(c*x + 1)) + a^3)/(c^2*x^2 - 1), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin((-c*x+1)**(1/2)/(c*x+1)**(1/2)))**3/(-c**2*x**2+1),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3/(-c^2*x^2+1),x, algorithm="giac")

[Out]

integrate(-(b*arcsin(sqrt(-c*x + 1)/sqrt(c*x + 1)) + a)^3/(c^2*x^2 - 1), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int -\frac {{\left (a+b\,\mathrm {asin}\left (\frac {\sqrt {1-c\,x}}{\sqrt {c\,x+1}}\right )\right )}^3}{c^2\,x^2-1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a + b*asin((1 - c*x)^(1/2)/(c*x + 1)^(1/2)))^3/(c^2*x^2 - 1),x)

[Out]

int(-(a + b*asin((1 - c*x)^(1/2)/(c*x + 1)^(1/2)))^3/(c^2*x^2 - 1), x)

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